General > General Technical Chat
Veritasium "How Electricity Actually Works"
electrodacus:
--- Quote from: hamster_nz on May 09, 2022, 05:12:42 am ---
If you agree with the above (current flows into the left hand side of the capacitor, and that current is flowing out of the right hand side of the capacitor, as it is being charged and both current flows stop when capacitor is fully charged) then you agree that during that time energy was flowing through the capacitor.
FTFY.
--- End quote ---
I assure you that energy is flowing in to capacitor and that distinction is very important.
It is irrelevant what else is in series with the capacitor and of course that thing in this case a resistor will see that charge current.
In case of Derek's experiment those first 65ns where used to charge the long transmission line and that charge current passed trough the lamp/resistor then after that the next 65ns the energy from the transmission line was discharged also in the lamp so nothing was lost. After those first 65ns the electron wave got to lamp through wires and continued to supply all the energy.
At no point in time energy was flowing to lamp/resistor outside the wires. Each unit of energy no matter how insignificant was delivered through wires and the wire resistance heated up from caring that energy.
You can not charge a capacitor by just connecting one of the terminals. Charging means moving electrons from one side of the capacitor to the other creating an excess of free electrons on one side and a deficit on the other side.
The dielectric can be anything including air or even vacuum and that is not what has the stored energy. The energy is contained in the capacitor plate same as transmission wires.
I'm sorry I fail at explaining things.
T3sl4co1l:
--- Quote from: electrodacus on May 09, 2022, 03:53:23 am ---I sort of get frustrated and annoyed of explaining what in/out and trough means.
--- End quote ---
So current is going into, but it's not? And it's going through, but it's not? And energy is going into, but it's also not?
It seems you can't produce a coherent description of any of the particulars in these matters, in addition to use of the above words... at least not in terms that anyone else can make any meaning of. If it's so frustrating and annoying, then... why bother?
You know the phrase, "better to be thought a fool..."?
Tim
hamster_nz:
--- Quote from: electrodacus on May 09, 2022, 06:29:15 am ---You can not charge a capacitor by just connecting one of the terminals.
--- End quote ---
If I were demonstrate to you an experiment which would charge a capacitor, without attaching any source to either terminal, and doesn't involve stupid things like firing ions at the terminals would that settle it for you?
To make things really clear, how about we have the capacitor as a 10cm^2 sheets of metal, separated by a suitable insulator?
The only other thing that is required is a removable switch, to discharge the capacitor when desired - say a screwdriver tip to short the two metal plates together, and of course some agreed way of measuring if the capacitor is charged or not. I suggest a standard DMM, that is used after the charging procedure?
Now two 10cm x 10cm metal sheets don't have much capacitance, so you will not get mA out of it. So don't expect wonders.. (heck this might be possible with just baking foil separated by tissues, I might give it a try!)
Are you game enough?
SandyCox:
--- Quote from: hamster_nz on May 09, 2022, 08:15:49 am ---
(heck this might be possible with just baking foil separated by tissues, I might give it a try!)
--- End quote ---
That is how I made capacitors when I was a child.
SandyCox:
Yes. This will work. Make a four-layer capacitor with foil and paper. Charge the outer two layers. You will then be able to extract energy from the capacitor formed by the inner two layers.
Navigation
[0] Message Index
[#] Next page
[*] Previous page
Go to full version